# Steps to finding a limit with squeeze theorem

How to find the limit of:

$$\lim_{n\to\infty} \sqrt[n]{n+\sin^2n}$$

using squeeze theorem?
Because $$0\le \sin^2n \le 1$$, I find $$a_n=\sqrt[n]{n}$$ (which is equal to 1) and $$c_n=\sqrt[n]{n+1}$$, but I don't know how to prove that second formula is also a 1. Could someone help me solve it please? Thank you!

• How do you know $\lim \sqrt[n]{n}=n$? You can't just assume. The proof of $\lim \sqrt[n]{n+1}$ isn't much different. – fleablood Feb 6 '20 at 19:32
• If you know that $2^{1/n} \rightarrow 1$ and $n^{1/n} \rightarrow 1,$ then you can use the fact that $\sqrt[n]{n} \leq \sqrt[n]{n + {\sin}^2 n} \leq \sqrt[n]{n+n} = 2^{1/n}n^{1/n},$ and note that the left side approaches $1$ and the right side approaches $1 \cdot 1 = 1.$ – Dave L. Renfro Feb 6 '20 at 19:34
• $n^{1/n} \lt (n+1)^{1/n} \lt (2n)^{1/n}=2^{1/n}n^{1/n}.$ – Peter Szilas Feb 6 '20 at 19:34
• Thank you both! Easier than I thought – L.krez Feb 6 '20 at 19:42

## 1 Answer

Here's a hint:

$$n

• (+1) Exactly what I gave in a comment, but your answer came first! – Dave L. Renfro Feb 6 '20 at 19:35
• Thank you, now it seems obvious! – L.krez Feb 6 '20 at 19:42